Discoveries about INFINITE SERIES

By Katy, 10th grader

BACKGROUND

Mr. Cohen had previously shown me an interesting property of infinite series. He started by having me find the sum of the infinite series ' + ' + 1/8 + 1/16 + 1/32 ' He suggested that I use a picture and color in a half, and then add a fourth and then add an eighth, a sixteenth, etc. until I found what the infinite series approached

The picture showed that the infinite series approaches 1.

Mr. Cohen next gave me the problem 1/3 + 1/9 + 1/27' This problem is a little trickier to show using diagrams, it's easier to use decimals (in Excel) to explain this problem:

1/3 |
.3333333333333 |

1/3 + 1/9 |
.4444444444444 |

1/3 + 1/9 + 1/27 |
.4814814814814 |

1/3 + 1/9 + 1/27 + 1/81 |
.493872 |

1/3 + 1/9 + 1/27 + 1/81 + 1/243 |
.497972 |

As you can see, this chart shows that the infinite series approaches .5 or ' . No matter how many terms you add, the series will not reach .5. It would only reach point five if you could keep adding into infinity.

The next problem was ' + 1/16 + 1/64 + 1/256'

(') |
0.25000000000 |

(1/4) + (1/16) |
0.312500000000 |

(1/4) + (1/16) + (1/64) |
0.328125000000 |

(1/4) + (1/16) + (1/64) + (1/256) |
0.332031250000 |

(1/4) + (1/16) + (1/64) + (1/256) + (1/1024) |
0.333007812500 |

(1/4) + (1/16) + (1/64) + (1/256) + (1/1024) + (1/4096) |
0.333251953 |

As you can see, each time another term is added, the infinite series gets closer to .333' or one third.

Mr. Cohen then asked me to figure a pattern for 1/n + 1/n^2 + 1/n^3 + 1/n^4'..

' + ' + 1/8 + 1/16 + 1/32 ' approached 1/1

1/3 + 1/9 + 1/27' approached 1/2

' + 1/16 + 1/64 + 1/256' approached 1/3

The rule seemed to be that **1/n + 1/n^2 + 1/n^3 +
1/n^4'.. will approach the limit 1/(n-1).**

In Mr. Cohen's book *Calculus by
and for Young People '--Worksheets *one of the problems caught my interest.
It involved **sharing two cookies between three people** using 'special
scissors' -- scissors that could only cut the cookies into two equal pieces. If
the special scissors were used once on each cookie, then there would be 4 pieces
and each person would get one piece, or ' a cookie. The remaining half-cookie
would be cut with the scissors into two pieces that would each be one fourth.
Since three people cannot evenly share two pieces they would not receive any of
the newly cut pieces. Another way to put it would be that each has person
received ' + 0/4 so far. The two left over pieces (which were each one fourth)
would be cut again, so there would be four pieces that were each a eighth of the
whole. Each person would get one of the new pieces, so they would have ' + 0/4
+ 1/8. The remaining eighth would be cut again. The two new pieces would each be
a sixteenth of the whole. Since two pieces cannot be divided between three
people, each person would not receive either of the new pieces. That means that
each person receives ' + 0/4 + 1/8 + 0/16. This pattern would continue, with the
cookie being cut forever and each person receiving **' + 0/4 + 1/8 + 0/16'**

Another way to write **' + 0/4 +
1/8 + 0/16 + 1/32'**

is ' + 1/8 + 1/32'

OR it can be written as 1/2 + 1/(2^3) + (1/2^5) ' I wondered if there was a rule that said what 1/n + 1/(n^3) + 1/(n^5)' , the sum of the odd powers, approached.

I already knew that ' + 1/(2^3) + (1/2^5) ' approached 2/3, so I next I tried to see what 1/3 + 1/(3^3) + 1/(3^5)' would approach.

1/3 |
0.3333333333 |

(1/3) + (1/27) |
0.3703703704 |

(1/3) + (1/27) + (1/243) |
0.3744855967 |

(1/3) + (1/27) + (1/243) + (1/2187) |
0.3749428441 |

(1/3) + (1/27) + (1/243) + (1/2187) + (1/19683) |
0.3749936493 |

As the table shows
1/3 + 1/(3^3) + 1/(3^5)' approaches .375 or
**3/8**

** **

** **

1/4 |
0.250000000 |

(1/4) + (1/64) |
0.265625000 |

(1/4) + (1/64) + (1/1024) |
0.266601563 |

(1/4) + (1/64) + (1/1024) + (1/16384) |
0.266662598 |

(1/4) + (1/64) + (1/1024) + (1/16384) + (1/262144) |
0.266666412 |

It approaches .26666666666666666666... or 4/15

** **

** **

So far, I'd learned:

' + 1/(2^3) + (1/2^5) ' approaches 2/3

1/3 + 1/(3^3) + 1/(3^5)' approaches 3/8

1/4 + 1/(4^3) + 1/(4^5)' approaches 4/15

This suggested the series **1/n + 1/(n^3) + 1/(n^5)'
approaches the limit n/(n^2 ' 1)**

** **

Since I'd found that

1/n + 1/n^2 + 1/n^3 + 1/n^4'.. approaches 1/n-1 and

1/n + 1/(n^3) + 1/(9^5)'approaches ** **n/(n^2 ' 1)

I wanted to find a rule for** **the limit of** 1/n + 1/(n^4) + 1/(n^7) +
1/(n^10)+...**

** **

I guessed that it would be n/(n^3 '1)

I tested my rule on the series ' + 1/16 + 1/128 '

1/2 |
0.5000000 |

1/2 +1/16 |
0.5625000 |

1/2 +1/16 + 1/128 |
0.5703125 |

1/2 +1/16 + 1/128 + 1/1024 |
0.5712891 |

1/2 +1/16 + 1/128 + 1/1024 +1/8192 |
0.5714111 |

1/2 +1/16 + 1/128 + 1/1024 +1/8192 + 1/65536 |
0.5714264 |

I thought it would approach 2/(2^3 ' 1) or 2/7, but 2/7 is .2857142. I used the computer to find out what fraction the infinite series was approaching, and it gave me 4/7, or .5714285

My new guess for the rule for **1/n + 1/(n^4) + 1/(n^7) +
1/(n^10) +... **was** (n^2)/(n^3 ' 1).** I tested this idea on 1/3 + 1/81 + 1/21897'

I thought that it would approach (3^2)/(3^3 ' 1) or 9/(27-1) or 9/26 or .3461538

When I added the first six terms of the series

1/3 |
0.33333333 |

1/3 +1/81 |
0.34567901 |

1/3 +1/81 + 1/2187 |
0.34613626 |

1/3 +1/81 + 1/2187 + 1/59049 |
0.34615319 |

1/3 +1/81 + 1/2187+ 1/59049 +1/1594323 |
0.34615382 |

1/3 +1/81 + 1/2187 + 1/59049 +1/1594323 + 1/43046721 |
0.34615385 |

I found that my guess was correct.

I took a look at the three rules I'd learned so far:

1/n + 1/n^2 + 1/n^3 + 1/n^4'.. approaches 1/n-1

1/n + 1/(n^3) +
1/(n^5)'approaches ** **n/(n^2 ' 1)

** **

1/n + 1/(n^4) + 1/(n^7) + 1/(n^10)' approaches (n^2)/(n^3 ' 1)

I noticed a very definite pattern. In the first series the numbers in the denominator did not skip any power as they went up (n^1, n^2, n^3'.) and the rule was (n^0)/(n^1 ' 1).

In the second series the numbers in the denominator skipped
one power each time (n^1, *skip n^2, *n^3, *skip n^4, *n^5, *skip
n^6*'*) *and the rule was (n^1)/(n^2 ' 1)

In the third series the numbers in the denominator skipped
two powers each time (n^1, *skip n^2 and n^3, *n^4, *skip n^5 and n^6,
*n^7') and the rule is (n^2)/(n^3 '1)

This
shows that a general rule is that if you have an infinite series with 1 in the
numerator, and y is the number of powers skipped in the denominator; then the
rule for the limit of that series will be **
(n^y)/(n^(y+1) - 1)**.

**Fine work
Katy!**

To download Don's materials

Mathman home